Câu hỏi 29

\(\begin{array}{l}\,\,\,\,\,\,\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {4{x^2} + 8x + 1}  + 2x} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\left( {\sqrt {4{x^2} + 8x + 1}  + 2x} \right)\left( {\sqrt {4{x^2} + 8x + 1}  - 2x} \right)}}{{\sqrt {4{x^2} + 8x + 1}  - 2x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{8x + 1}}{{\sqrt {4{x^2} + 8x + 1}  - 2x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{8 + \dfrac{1}{x}}}{{ - \sqrt {4 + \dfrac{8}{x} + \dfrac{1}{{{x^2}}}}  - 2}} = \dfrac{8}{{ - 2 - 2}} =  - 2.\end{array}\)

Chọn: C